How to Use Calculus to Rotate Curves Around an Axis

Start by opening a new workbook in Excel from the desktop, from the dock, or from within your Applications folder inside the Microsoft folder.,In Preferences, set R1C1 to unchecked or Off, set Ribbon to checked or On and set Show Formula Bar to...

12 Steps 4 min read Advanced

Step-by-Step Guide

  1. Step 1: Start by opening a new workbook in Excel from the desktop

    Double click on Excel (either the green X on the dock or the app title in the folder) and select File New Workbook. ,, Doing so will select the entire worksheet.

    Format Cells Number Number to decimal places 2, show comma.

    Format Cells Alignment Center. # Title the first worksheet, "Rotate Function f(x)" and save the workbook as "Rotate Curves About An Axis" into an appropriate folder such as 'Microsoft Excel Imagery' or 'LifeGuide Hub Articles'., You want to define the volume of the solid of revolution generated by revolving about the x-axis the region R which is bounded by the curve y = f(x), the x-axis, and the vertical lines x = a and x = b.

    Let f(x) = sqrt(x) and a = 1 and b =
    4.

    Subdivide the interval into n subintervals by a partition P, and choose n points wi, one in each subinterval.

    Draw n approximating rectangles with base and altitude f(wi), i = 1, 2, 3, ... , n; a typical one of these rectangles is shown in the diagram as Rect HGFE.

    Revolve the region R about the x-axis to generate a solid of revolution, using the n rectangles to sweep out n right circular cylinders.

    The cylinders swept out by the typical rectangle, eg.

    Rect HGFE, is shown in the following Diagram; since the radius of its base is f(wi) and its altitude is ∆xi, its volume is ∆Vi = π*^2 *∆xi.

    Note that if you want to create a washer type of form, the formula changes to π * ∫ba *dx
    -- so it is a definite integral of the difference of the squares of the external function, f(x), and the internal (hole) function, g(x).

    Note also that you may let f be a continuous function on and if the region bounded by y = f(x), the x-axis, and the lines x = a and x = b lies in the first quadrant, the volume of the solid of revolution generated by revolving this region about the y-axis is V = 2π * ∫ba x*f(x)*dx, another definite integral. , If the arc of the curve y = f(x), from the point (a, f(a)) to the point (b, f(b)) is revolved about the x-axis, a surface of revolution S is swept out.

    Find the area of the surface of revolution by first partitioning into n intervals , i = 1, 2, 3, ..., n.

    Let Qi be the point on the curve whose coordinates are (xi,f(xi)), and denote the point (a, f(a)) by Q0.

    Then let the broken line formed by the n chords Qi-1Qi of the curve be revolved about the x-axis; it sweeps out a surface which approximates S, and this approximation improves as the norm |P| of the partition decreases.

    Consider that the lateral area of a frustum of a cone, having slant height s and radius of its bases r1 and r2, is π*(r1 + r2)*s.

    Thus each chord Qi-1Qi, as it revolves about the x-axis, sweeps out the lateral surface of a frustum of a cone whose area is π**|Qi-1*Qi|.

    Consider that, because of the formula for the arc distance (see the article Approximate Arc Length Using the Distance Formula), this may be re-written and defined as follows:
    Let f and f' be continuous on with f(x) ⩾ 0 for a ⩽ x ⩽ b.

    The area of the surface of revolution swept out by revolving about the x-axis the segment of the curve y = f(x), from the point (a, f(a)) to the point (b, f(b)) is: 2π * ∫ba f(x)*sqrt(1+f'(x)^2)*dx.

    Example:
    Find the area of the surface of revolution generated by revolving about the x-axis the segment of the curve y = sqrt(x) from (1,1) to (4,2).

    Solution:
    By substituting f(x) = sqrt(x) and f '(x) = 1/(2*sqrt(x)) in the above formula, you get: 2π * ∫41 x^.5 * sqrt(1+(1/(2*sqrt(x)))^2)*dx = π * ∫41 sqrt(4x +1) dx (by dividing by sqrt(4) = π/4 * ∫41 (4x +1)^.5 * d(4x +1) = π/4 * /(3/2)41 (by integration) = π/4 * 2/3 * (17^1.5
    - 5^1.5) = π/6 * (17^1.5
    - 5^1.5) =
    30.8465 √
  2. Step 2: from the dock

  3. Step 3: or from within your Applications folder inside the Microsoft folder.

  4. Step 4: In Preferences

  5. Step 5: set R1C1 to unchecked or Off

  6. Step 6: set Ribbon to checked or On and set Show Formula Bar to checked or On.

  7. Step 7: Click in the far upper left top corner above the 1 of row 1 and to the left of column A.

  8. Step 8: Enter to cell A1 the following text and then set Format Cell Alignment to Wrap Text: Let f be a function which is continuous on the closed interval

  9. Step 9: with f(x) ≥ 0 for a ≤ x ≤ b.

  10. Step 10: Consider a function f which is continuous on the interval

  11. Step 11: with f(x) ⊵ 0 for a ⊴ x ⊴ b

  12. Step 12: and whose first derivative f' is also continuous on .

Detailed Guide

Double click on Excel (either the green X on the dock or the app title in the folder) and select File New Workbook. ,, Doing so will select the entire worksheet.

Format Cells Number Number to decimal places 2, show comma.

Format Cells Alignment Center. # Title the first worksheet, "Rotate Function f(x)" and save the workbook as "Rotate Curves About An Axis" into an appropriate folder such as 'Microsoft Excel Imagery' or 'LifeGuide Hub Articles'., You want to define the volume of the solid of revolution generated by revolving about the x-axis the region R which is bounded by the curve y = f(x), the x-axis, and the vertical lines x = a and x = b.

Let f(x) = sqrt(x) and a = 1 and b =
4.

Subdivide the interval into n subintervals by a partition P, and choose n points wi, one in each subinterval.

Draw n approximating rectangles with base and altitude f(wi), i = 1, 2, 3, ... , n; a typical one of these rectangles is shown in the diagram as Rect HGFE.

Revolve the region R about the x-axis to generate a solid of revolution, using the n rectangles to sweep out n right circular cylinders.

The cylinders swept out by the typical rectangle, eg.

Rect HGFE, is shown in the following Diagram; since the radius of its base is f(wi) and its altitude is ∆xi, its volume is ∆Vi = π*^2 *∆xi.

Note that if you want to create a washer type of form, the formula changes to π * ∫ba *dx
-- so it is a definite integral of the difference of the squares of the external function, f(x), and the internal (hole) function, g(x).

Note also that you may let f be a continuous function on and if the region bounded by y = f(x), the x-axis, and the lines x = a and x = b lies in the first quadrant, the volume of the solid of revolution generated by revolving this region about the y-axis is V = 2π * ∫ba x*f(x)*dx, another definite integral. , If the arc of the curve y = f(x), from the point (a, f(a)) to the point (b, f(b)) is revolved about the x-axis, a surface of revolution S is swept out.

Find the area of the surface of revolution by first partitioning into n intervals , i = 1, 2, 3, ..., n.

Let Qi be the point on the curve whose coordinates are (xi,f(xi)), and denote the point (a, f(a)) by Q0.

Then let the broken line formed by the n chords Qi-1Qi of the curve be revolved about the x-axis; it sweeps out a surface which approximates S, and this approximation improves as the norm |P| of the partition decreases.

Consider that the lateral area of a frustum of a cone, having slant height s and radius of its bases r1 and r2, is π*(r1 + r2)*s.

Thus each chord Qi-1Qi, as it revolves about the x-axis, sweeps out the lateral surface of a frustum of a cone whose area is π**|Qi-1*Qi|.

Consider that, because of the formula for the arc distance (see the article Approximate Arc Length Using the Distance Formula), this may be re-written and defined as follows:
Let f and f' be continuous on with f(x) ⩾ 0 for a ⩽ x ⩽ b.

The area of the surface of revolution swept out by revolving about the x-axis the segment of the curve y = f(x), from the point (a, f(a)) to the point (b, f(b)) is: 2π * ∫ba f(x)*sqrt(1+f'(x)^2)*dx.

Example:
Find the area of the surface of revolution generated by revolving about the x-axis the segment of the curve y = sqrt(x) from (1,1) to (4,2).

Solution:
By substituting f(x) = sqrt(x) and f '(x) = 1/(2*sqrt(x)) in the above formula, you get: 2π * ∫41 x^.5 * sqrt(1+(1/(2*sqrt(x)))^2)*dx = π * ∫41 sqrt(4x +1) dx (by dividing by sqrt(4) = π/4 * ∫41 (4x +1)^.5 * d(4x +1) = π/4 * /(3/2)41 (by integration) = π/4 * 2/3 * (17^1.5
- 5^1.5) = π/6 * (17^1.5
- 5^1.5) =
30.8465 √

About the Author

K

Kenneth Wood

Creates helpful guides on crafts to inspire and educate readers.

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